Problem 88 - College Algebra and Calculus: An Applied Approach

Question

Find the standard form of the equation of the specified circle. Center: \(\left(0, \frac{1}{2}\right)\); radius: \(\frac{2}{3}\)

Step-by-Step Solution

Answer

The standard form equation of the circle is \( x^2 + (y - \frac{1}{2})^2 = \frac{4}{9} \)

Step 1: Substituting the given values

Substitute the given values for the center and the radius into the formula. Here, a = 0, b = 1/2, and r = 2/3.

Step 2: Implementation of formula

The formula is then implemented as follows: \( (x-0)^2 + (y- \frac{1}{2})^2 = \left(\frac{2}{3}\right)^2 \)

Step 3: Simplifying the equation

Simplify the equation by squaring the terms and simplifying the fractions. The final equation obtained is \( x^2 + (y - \frac{1}{2})^2 = \frac{4}{9} \)

Key Concepts

Circle Equations

In algebra and geometry, a circle can be defined mathematically as a set of all points in a plane that are at a fixed distance, known as the radius, from a given point called the center. When we deal with circle equations, especially in the Cartesian coordinate system, we typically use the standard form equation:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
Here, \( (a, b) \) denotes the center of the circle, and \( r \) is the radius. To fully understand this concept, it's essential to visualize a circle in a coordinate plane and observe that regardless of the direction from the center, the distance of the circumference from the center is always equal, which is why circles are symmetric and the equation is consistent for all points on the circle.

In the case of the given exercise, the center of the circle is \( (0, \frac{1}{2}) \) and the radius is \( \frac{2}{3} \). This implies that the standard form of the equation includes substituting 0 for \( a \) and \( \frac{1}{2} \) for \( b \), with the radius squared becoming \( \frac{4}{9} \). Thus, after substitution and squaring the radius, the equation takes the shape of the one derived in the exercise.

Algebraic Operations

The process of finding the standard form of the equation of a circle involves several algebraic operations. This includes simple substitutions, squaring numbers, expanding binomials, and simplifying fractions.

Squaring Binomials

Squaring a binomial such as \( (x-a)^2 \) or \( (y-b)^2 \) results in \( x^2 - 2ax + a^2 \) and \( y^2 - 2by + b^2 \) respectively. In our exercise, simplifying \( (x-0)^2 \) is straightforward as it becomes \( x^2 \), and \( (y- \frac{1}{2})^2 \) requires squaring both the term \( y \) and the constant \( \frac{1}{2} \).

Simplifying Fractions

Furthermore, simplifying fractions such as squaring \( \frac{2}{3} \) to obtain \( \frac{4}{9} \) is another algebraic operation. Understanding how to manipulate these expressions and perform these operations correctly is crucial to solve and understand equations representing geometric figures in algebra.

Cartesian Coordinates System

The Cartesian coordinates system is a 2D system used for graphing points, lines, and curves, including circles. It consists of two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical). Points are located using an ordered pair \( (x, y) \), which represents their position relative to these axes.

For the circle, the center's coordinates \( (a, b) \) place it uniquely on the plane. In the given problem, the center at \( (0, \frac{1}{2}) \) suggests it is on the y-axis \( \frac{1}{2} \) units above the origin. Every point on the circle is a fixed radius away from this center. By understanding the Cartesian coordinate system, one can visualize where the circle will be placed and how its standard equation represents every point lying on its boundary.